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Chapter 3 Practice Problems (Practice – no credit) Due: 11:59pm on Wednesday, February 12, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Tactics Box 3.1 Determining the Components of a Vector Learning Goal: To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector is decomposed into component vectors and parallel to the coordinate axes, we can describe each component vector with a single number (a scalar) called the component. This tactics box describes how to determine the x component and y component of vector , denoted and . TACTICS BOX 3.1 Determining the components of a vector The absolute value of the x component is the magnitude of the 1. component vector . 2. The sign of is positive if points in the positive x direction; it is negative if points in the negative x direction. 3. The y component is determined similarly. Part A What is the magnitude of the component vector shown in the figure? Express your answer in meters to one significant figure. A A x A y A Ax Ay |Ax| Ax A x Ax A x A x Ay A x ANSWER: Answer Requested Part B What is the sign of the y component of vector shown in the figure? ANSWER: Correct Part C Now, combine the information given in the tactics box above to find the x and y components, and , of vector shown in the figure. |Ax| = 5 m Ay A positive negative Bx By B Express your answers, separated by a comma, in meters to one significant figure. ANSWER: Correct Vector Components–Review Learning Goal: To introduce you to vectors and the use of sine and cosine for a triangle when resolving components. Vectors are an important part of the language of science, mathematics, and engineering. They are used to discuss multivariable calculus, electrical circuits with oscillating currents, stress and strain in structures and materials, and flows of atmospheres and fluids, and they have many other applications. Resolving a vector into components is a precursor to computing things with or about a vector quantity. Because position, velocity, acceleration, force, momentum, and angular momentum are all vector quantities, resolving vectors into components is the most important skill required in a mechanics course. The figure shows the components of , and , along the x and y axes of the coordinate system, respectively. The components of a vector depend on the coordinate system’s orientation, the key being the angle between the vector and the coordinate axes, often designated . Bx, By = -2,-5 m, m F Fx Fy Part A The figure shows the standard way of measuring the angle. is measured to the vector from the x axis, and counterclockwise is positive. Express and in terms of the length of the vector and the angle , with the components separated by a comma. ANSWER: Fx Fy F Fx, Fy = Fcos, Fsin Correct In principle, you can determine the components of any vector with these expressions. If lies in one of the other quadrants of the plane, will be an angle larger than 90 degrees (or in radians) and and will have the appropriate signs and values. Unfortunately this way of representing , though mathematically correct, leads to equations that must be simplified using trig identities such as and . These must be used to reduce all trig functions present in your equations to either or . Unless you perform this followup step flawlessly, you will fail to recoginze that , and your equations will not simplify so that you can progress further toward a solution. Therefore, it is best to express all components in terms of either or , with between 0 and 90 degrees (or 0 and in radians), and determine the signs of the trig functions by knowing in which quadrant the vector lies. Part B When you resolve a vector into components, the components must have the form or . The signs depend on which quadrant the vector lies in, and there will be one component with and the other with . In real problems the optimal coordinate system is often rotated so that the x axis is not horizontal. Furthermore, most vectors will not lie in the first quadrant. To assign the sine and cosine correctly for vectors at arbitrary angles, you must figure out which angle is and then properly reorient the definitional triangle. As an example, consider the vector shown in the diagram labeled “tilted axes,” where you know the angle between and the y axis. Which of the various ways of orienting the definitional triangle must be used to resolve into components in the tilted coordinate system shown? (In the figures, the hypotenuse is orange, the side adjacent to is red, and the side opposite is yellow.) F /2 cos() sin() F sin(180 + ) = −sin() cos(90 + ) = −sin() sin() cos() sin(180 + ) + cos(270 − ) = 0 sin() cos() /2 F |F| cos() |F| sin() sin() cos() N N N Indicate the number of the figure with the correct orientation. Hint 1. Recommended procedure for resolving a vector into components First figure out the sines and cosines of , then figure out the signs from the quadrant the vector is in and write in the signs. Hint 2. Finding the trigonometric functions Sine and cosine are defined according to the following convention, with the key lengths shown in green: The hypotenuse has unit length, the side adjacent to has length , and the cos() side opposite has length . The colors are chosen to remind you that the vector sum of the two orthogonal sides is the vector whose magnitude is the hypotenuse; red + yellow = orange. ANSWER: Correct Part C Choose the correct procedure for determining the components of a vector in a given coordinate system from this list: ANSWER: sin() 1 2 3 4 Correct Part D The space around a coordinate system is conventionally divided into four numbered quadrants depending on the signs of the x and y coordinates . Consider the following conditions: A. , B. , C. , D. , Which of these lettered conditions are true in which the numbered quadrants shown in ? Write the answer in the following way: If A were true in the third quadrant, B in the second, C in the first, and D in the fourth, enter “3, 2, 1, 4” as your response. ANSWER: Align the adjacent side of a right triangle with the vector and the hypotenuse along a coordinate direction with as the included angle. Align the hypotenuse of a right triangle with the vector and an adjacent side along a coordinate direction with as the included angle. Align the opposite side of a right triangle with the vector and the hypotenuse along a coordinate direction with as the included angle. Align the hypotenuse of a right triangle with the vector and the opposite side along a coordinate direction with as the included angle. x > 0 y > 0 x > 0 y < 0 x < 0 y > 0 x < 0 y < 0 Correct Part E Now find the components and of in the tilted coordinate system of Part B. Express your answer in terms of the length of the vector and the angle , with the components separated by a comma. ANSWER: Answer Requested ± Resolving Vector Components with Trigonometry Often a vector is specified by a magnitude and a direction; for example, a rope with tension exerts a force of magnitude in a direction 35 north of east. This is a good way to think of vectors; however, to calculate results with vectors, it is best to select a coordinate system and manipulate the components of the vectors in that coordinate system. Nx Ny N N Nx, Ny = −Nsin(),Ncos() T T Part A Find the components of the vector with length = 1.00 and angle =10.0 with respect to the x axis as shown. Enter the x component followed by the y component, separated by a comma. Hint 1. What is the x component? Look at the figure shown. points in the positive x direction, so is positive. Also, the magnitude is just the length . ANSWER: Correct Part B Find the components of the vector with length = 1.00 and angle =15.0 with respect to the x axis as shown. Enter the x component followed by the y component, separated by a comma. A a A x Ax |Ax| OL = OMcos() A = 0.985,0.174 B b Hint 1. What is the x component? The x component is still of the same form, that is, . ANSWER: Correct The components of still have the same form, that is, , despite 's placement with respect to the y axis on the drawing. Part C Find the components of the vector with length = 1.00 and angle 35.0 as shown. Enter the x component followed by the y component, separated by a comma. Hint 1. Method 1: Find the angle that makes with the positive x axis Angle = 0.611 differs from the other two angles because it is the angle between the vector and the y axis, unlike the others, which are with respect to the x axis. What is the angle that makes with the positive x axis? Express your answer numerically in degrees. ANSWER: Hint 2. Method 2: Use vector addition Look at the figure shown. Lcos() B = 0.966,0.259 B (Lcos(), Lsin()) B C c = C C 125 1. . 2. . 3. , the x component of is negative, since points in the negative x direction. Use this information to find . Similarly, find . ANSWER: Answer Requested ± Vector Addition and Subtraction In general it is best to conceptualize vectors as arrows in space, and then to make calculations with them using their components. (You must first specify a coordinate system in order to find the components of each arrow.) This problem gives you some practice with the components. Let vectors , , and . Calculate the following, and express your answers as ordered triplets of values separated by commas. Part A ANSWER: Correct C = C + x C y |C | = length(QR) = c sin() x Cx C C x Cx Cy C = -0.574,0.819 A = (1, 0,−3) B = (−2, 5, 1) C = (3, 1, 1) A − B = 3,-5,-4 Part B ANSWER: Correct Part C ANSWER: Correct Part D ANSWER: Correct B − C = -5,4,0 −A + B − C = -6,4,3 3A − 2C = -3,-2,-11 Part E ANSWER: Correct Part F ANSWER: Correct Video Tutor: Balls Take High and Low Tracks First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the questions at right. You can watch the video again at any point. Part A −2A + 3B − C = -11,14,8 2A − 3(B − C) = 17,-12,-6 Consider the video demonstration that you just watched. Which of the following changes could potentially allow the ball on the straight inclined (yellow) track to win? Ignore air resistance. Select all that apply. Hint 1. How to approach the problem Answers A and B involve changing the steepness of part or all of the track. Answers C and D involve changing the mass of the balls. So, first you should decide which of those factors, if either, can change how fast the ball gets to the end of the track. ANSWER: Correct If the yellow track were tilted steeply enough, its ball could win. How might you go about calculating the necessary change in tilt? Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. A. Increase the tilt of the yellow track. B. Make the downhill and uphill inclines on the red track less steep, while keeping the total distance traveled by the ball the same. C. Increase the mass of the ball on the yellow track. D. Decrease the mass of the ball on the red track.

Chapter 1 Practice Problems (Practice – no credit) Due: 11:59pm on Wednesday, February 5, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Curved Motion Diagram The motion diagram shown in the figure represents a pendulum released from rest at an angle of 45 from the vertical. The dots in the motion diagram represent the positions of the pendulum bob at eleven moments separated by equal time intervals. The green arrows represent the average velocity between adjacent dots. Also given is a “compass rose” in which directions are labeled with the letters of the alphabet. Part A What is the direction of the acceleration of the object at moment 5? Enter the letter of the arrow with this direction from the compass rose in the figure. Type Z if the acceleration vector has zero length. You did not open hints for this part. ANSWER: Incorrect; Try Again Part B What is the direction of the acceleration of the object at moments 0 and 10? Enter the letters corresponding to the arrows with these directions from the compass rose in the figure, separated by commas. Type Z if the acceleration vector has zero length. You did not open hints for this part. ANSWER: Incorrect; Try Again PSS 1.1 Motion Diagrams Learning Goal: To practice Problem-Solving Strategy 1.1 for motion diagram problems. A car is traveling with constant velocity along a highway. The driver notices he is late for work, so he stomps down on the gas pedal and the car begins to speed up. The car has just achieved double its directions at time step 0, time step 10 = initial velocity when the driver spots a police officer behind him and applies the brakes. The car then slows down, coming to rest at a stoplight ahead. Draw a complete motion diagram for this situation. PROBLEM-SOLVING STRATEGY 1.1 Motion diagrams MODEL: Represent the moving object as a particle. Make simplifying assumptions when interpreting the problem statement. VISUALIZE: A complete motion diagram consists of: The position of the object in each frame of the film, shown as a dot. Use five or six dots to make the motion clear but without overcrowding the picture. More complex motions may need more dots. The average velocity vectors, found by connecting each dot in the motion diagram to the next with a vector arrow. There is one velocity vector linking each set of two position dots. Label the row of velocity vectors . The average acceleration vectors, found using Tactics Box 1.3. There is one acceleration vector linking each set of two velocity vectors. Each acceleration vector is drawn at the dot between the two velocity vectors it links. Use to indicate a point at which the acceleration is zero. Label the row of acceleration vectors . Model It is appropriate to use the particle model for the car. You should also make some simplifying assumptions. v 0 a Part A The car’s motion can be divided into three different stages: its motion before the driver realizes he’s late, its motion after the driver hits the gas (but before he sees the police car), and its motion after the driver sees the police car. Which of the following simplifying assumptions is it reasonable to make in this problem? During each of the three different stages of its motion, the car is moving with constant A. acceleration. B. During each of the three different stages of its motion, the car is moving with constant velocity. C. The highway is straight (i.e., there are no curves). D. The highway is level (i.e., there are no hills or valleys). Enter all the correct answers in alphabetical order without commas. For example, if statements C and D are correct, enter CD. ANSWER: Correct In addition to the assumptions listed above, in the rest of this problem assume that the car is moving in a straight line to the right. Visualize Part B In the three diagrams shown to the left, the position of the car at five subsequent instants of time is represented by black dots, and the car’s average velocity is represented by green arrows. Which of these diagrams best describes the position and the velocity of the car before the driver notices he is late? ANSWER: Correct Part C Which of the diagrams shown to the left best describes the position and the velocity of the car after the driver hits the gas, but before he notices the police officer? ANSWER: Correct A B C A B C Part D Which of the diagrams shown to the left best describes the position and the velocity of the car after the driver notices the police officer? ANSWER: Correct Part E Which of the diagrams shown below most accurately depicts the average acceleration vectors of the car during the events described in the problem introduction? ANSWER: A B C Correct You can now draw a complete motion diagram for the situation described in this problem. Your diagram should look like this: Measurements in SI Units Familiarity with SI units will aid your study of physics and all other sciences. Part A What is the approximate height of the average adult in centimeters? Hint 1. Converting between feet and centimeters The distance from your elbow to your fingertips is typically about 50 . A B C cm ANSWER: Correct If you’re not familiar with metric units of length, you can use your body to develop intuition for them. The average height of an adult is 5 6.4 . The distance from elbow to fingertips on the average adult is about 50 . Ten (1 ) is about the width of this adult’s little finger and 10 is about the width of the average hand. Part B Approximately what is the mass of the average adult in kilograms? Hint 1. Converting between pounds and kilograms Something that weighs 1 has a mass of about . ANSWER: Correct Something that weighs 1 has a mass of about . This is a useful conversion to keep in mind! ± A Trip to Europe 100 200 300 cm cm cm feet inches cm mm cm cm pound 1 kg 2 80 500 1200 kg kg kg pound (1/2) kg Learning Goal: To understand how to use dimensional analysis to solve problems. Dimensional analysis is a useful tool for solving problems that involve unit conversions. Since unit conversion is not limited to physics problems but is part of our everyday life, correct use of conversion factors is essential to working through problems of practical importance. For example, dimensional analysis could be used in problems involving currency exchange. Say you want to calculate how many euros you get if you exchange 3600 ( ), given the exchange rate , that is, 1 to 1.20 . Begin by writing down the starting value, 3600 . This can also be written as a fraction: . Next, convert dollars to euros. This conversion involves multiplying by a simple conversion factor derived from the exchange rate: . Note that the “dollar” unit, , should appear on the bottom of this conversion factor, since appears on the top of the starting value. Finally, since dollars are divided by dollars, the units can be canceled and the final result is . Currency exchange is only one example of many practical situations where dimensional analysis may help you to work through problems. Remember that dimensional analysis involves multiplying a given value by a conversion factor, resulting in a value in the new units. The conversion factor can be the ratio of any two quantities, as long as the ratio is equal to one. You and your friends are organizing a trip to Europe. Your plan is to rent a car and drive through the major European capitals. By consulting a map you estimate that you will cover a total distance of 5000 . Consider the euro-dollar exchange rate given in the introduction and use dimensional analysis to work through these simple problems. Part A You select a rental package that includes a car with an average consumption of 6.00 of fuel per 100 . Considering that in Europe the average fuel cost is 1.063 , how much (in US dollars) will you spend in fuel on your trip? Express your answer numerically in US dollars to three significant figures. You did not open hints for this part. ANSWER: US dollars USD 1 EUR = 1.20 USD euro US dollars USD 3600 USD 1 1.00 EUR 1.20 USD USD USD ( )( ) = 3000 EUR 3600 USD 1 1.00 EUR 1.20 USD km liters km euros/liter Part B How many gallons of fuel would the rental car consume per mile? Express your answer numerically in gallons per mile to three significant figures. You did not open hints for this part. ANSWER: Part C This question will be shown after you complete previous question(s). Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. Cost of fuel = USD gallons/mile

Chapter 6 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, March 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy PSS 6.1 Equilibrium Problems Learning Goal: To practice Problem-Solving Strategy 6.1 for equilibrium problems. A pair of students are lifting a heavy trunk on move-in day. Using two ropes tied to a small ring at the center of the top of the trunk, they pull the trunk straight up at a constant velocity . Each rope makes an angle with respect to the vertical. The gravitational force acting on the trunk has magnitude . Find the tension in each rope. PROBLEM-SOLVING STRATEGY 6.1 Equilibrium problems MODEL: Make simplifying assumptions. VISUALIZE: Establish a coordinate system, define symbols, and identify what the problem is asking you to find. This is the process of translating words into symbols. Identify all forces acting on the object, and show them on a free-body diagram. These elements form the pictorial representation of the problem. SOLVE: The mathematical representation is based on Newton’s first law: . The vector sum of the forces is found directly from the free-body diagram. v FG T F = = net i F i 0 ASSESS: Check if your result has the correct units, is reasonable, and answers the question. Model The trunk is moving at a constant velocity. This means that you can model it as a particle in dynamic equilibrium and apply the strategy above. Furthermore, you can ignore the masses of the ropes and the ring because it is reasonable to assume that their combined weight is much less than the weight of the trunk. Visualize Part A The most convenient coordinate system for this problem is one in which the y axis is vertical and the ropes both lie in the xy plane, as shown below. Identify the forces acting on the trunk, and then draw a free-body diagram of the trunk in the diagram below. The black dot represents the trunk as it is lifted by the students. Draw the vectors starting at the black dot. The location and orientation of the vectors will be graded. The length of the vectors will not be graded. ANSWER: Part B This question will be shown after you complete previous question(s). Solve Part C This question will be shown after you complete previous question(s). Assess Part D This question will be shown after you complete previous question(s). A Gymnast on a Rope A gymnast of mass 70.0 hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume that the rope does not stretch. Use the value for the acceleration of gravity. Part A Calculate the tension in the rope if the gymnast hangs motionless on the rope. Express your answer in newtons. You did not open hints for this part. ANSWER: Part B Calculate the tension in the rope if the gymnast climbs the rope at a constant rate. Express your answer in newtons. You did not open hints for this part. kg 9.81m/s2 T T = N T ANSWER: Part C Calculate the tension in the rope if the gymnast climbs up the rope with an upward acceleration of magnitude 1.10 . Express your answer in newtons. You did not open hints for this part. ANSWER: Part D Calculate the tension in the rope if the gymnast slides down the rope with a downward acceleration of magnitude 1.10 . Express your answer in newtons. You did not open hints for this part. ANSWER: T = N T m/s2 T = N T m/s2 T = N Applying Newton’s 2nd Law Learning Goal: To learn a systematic approach to solving Newton’s 2nd law problems using a simple example. Once you have decided to solve a problem using Newton’s 2nd law, there are steps that will lead you to a solution. One such prescription is the following: Visualize the problem and identify special cases. Isolate each body and draw the forces acting on it. Choose a coordinate system for each body. Apply Newton’s 2nd law to each body. Write equations for the constraints and other given information. Solve the resulting equations symbolically. Check that your answer has the correct dimensions and satisfies special cases. If numbers are given in the problem, plug them in and check that the answer makes sense. Think about generalizations or simplfications of the problem. As an example, we will apply this procedure to find the acceleration of a block of mass that is pulled up a frictionless plane inclined at angle with respect to the horizontal by a perfect string that passes over a perfect pulley to a block of mass that is hanging vertically. Visualize the problem and identify special cases First examine the problem by drawing a picture and visualizing the motion. Apply Newton’s 2nd law, , to each body in your mind. Don’t worry about which quantities are given. Think about the forces on each body: How are these consistent with the direction of the acceleration for that body? Can you think of any special cases that you can solve quickly now and use to test your understanding later? m2 m1 F = ma One special case in this problem is if , in which case block 1 would simply fall freely under the acceleration of gravity: . Part A Consider another special case in which the inclined plane is vertical ( ). In this case, for what value of would the acceleration of the two blocks be equal to zero? Express your answer in terms of some or all of the variables and . ANSWER: Isolate each body and draw the forces acting on it A force diagram should include only real forces that act on the body and satisfy Newton’s 3rd law. One way to check if the forces are real is to detrmine whether they are part of a Newton’s 3rd law pair, that is, whether they result from a physical interaction that also causes an opposite force on some other body, which may not be part of the problem. Do not decompose the forces into components, and do not include resultant forces that are combinations of other real forces like centripetal force or fictitious forces like the “centrifugal” force. Assign each force a symbol, but don’t start to solve the problem at this point. Part B Which of the four drawings is a correct force diagram for this problem? = 0 m2 = −g a 1 j ^ = /2 m1 m2 g m1 = ANSWER: Choose a coordinate system for each body Newton’s 2nd law, , is a vector equation. To add or subtract vectors it is often easiest to decompose each vector into components. Whereas a particular set of vector components is only valid in a particular coordinate system, the vector equality holds in any coordinate system, giving you freedom to pick a coordinate system that most simplifies the equations that result from the component equations. It’s generally best to pick a coordinate system where the acceleration of the system lies directly on one of the coordinate axes. If there is no acceleration, then pick a coordinate system with as many unknowns as possible along the coordinate axes. Vectors that lie along the axes appear in only one of the equations for each component, rather than in two equations with trigonometric prefactors. Note that it is sometimes advantageous to use different coordinate systems for each body in the problem. In this problem, you should use Cartesian coordinates and your axes should be stationary with respect to the inclined plane. Part C Given the criteria just described, what orientation of the coordinate axes would be best to use in this problem? In the answer options, “tilted” means with the x axis oriented parallel to the plane (i.e., at angle to the horizontal), and “level” means with the x axis horizontal. ANSWER: Apply Newton’s 2nd law to each body a b c d F = ma tilted for both block 1 and block 2 tilted for block 1 and level for block 2 level for block 1 and tilted for block 2 level for both block 1 and block 2 Part D What is , the sum of the x components of the forces acting on block 2? Take forces acting up the incline to be positive. Express your answer in terms of some or all of the variables tension , , the magnitude of the acceleration of gravity , and . You did not open hints for this part. ANSWER: Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). Part G This question will be shown after you complete previous question(s). Lifting a Bucket A 6- bucket of water is being pulled straight up by a string at a constant speed. F2x T m2 g m2a2x =F2x = kg Part A What is the tension in the rope? ANSWER: Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Friction Force on a Dancer on a Drawbridge A dancer is standing on one leg on a drawbridge that is about to open. The coefficients of static and kinetic friction between the drawbridge and the dancer’s foot are and , respectively. represents the normal force exerted on the dancer by the bridge, and represents the gravitational force exerted on the dancer, as shown in the drawing . For all the questions, we can assume that the bridge is a perfectly flat surface and lacks the curvature characteristic of most bridges. about 42 about 60 about 78 0 because the bucket has no acceleration. N N N N μs μk n F g Part A Before the drawbridge starts to open, it is perfectly level with the ground. The dancer is standing still on one leg. What is the x component of the friction force, ? Express your answer in terms of some or all of the variables , , and/or . You did not open hints for this part. ANSWER: Part B The drawbridge then starts to rise and the dancer continues to stand on one leg. The drawbridge stops just at the point where the dancer is on the verge of slipping. What is the magnitude of the frictional force now? Express your answer in terms of some or all of the variables , , and/or . The angle should not appear in your answer. F f n μs μk Ff = Ff n μs μk You did not open hints for this part. ANSWER: Part C Then, because the bridge is old and poorly designed, it falls a little bit and then jerks. This causes the person to start to slide down the bridge at a constant speed. What is the magnitude of the frictional force now? Express your answer in terms of some or all of the variables , , and/or . The angle should not appear in your answer. ANSWER: Part D The bridge starts to come back down again. The dancer stops sliding. However, again because of the age and design of the bridge it never makes it all the way down; rather it stops half a meter short. This half a meter corresponds to an angle degree (see the diagram, which has the angle exaggerated). What is the force of friction now? Express your answer in terms of some or all of the variables , , and . Ff = Ff n μs μk Ff = 1 Ff n Fg You did not open hints for this part. ANSWER: Kinetic Friction Ranking Task Below are eight crates of different mass. The crates are attached to massless ropes, as indicated in the picture, where the ropes are marked by letters. Each crate is being pulled to the right at the same constant speed. The coefficient of kinetic friction between each crate and the surface on which it slides is the same for all eight crates. Ff = Part A Rank the ropes on the basis of the force each exerts on the crate immediately to its left. Rank from largest to smallest. To rank items as equivalent, overlap them. You did not open hints for this part. ANSWER: Pushing a Block Learning Goal: To understand kinetic and static friction. A block of mass lies on a horizontal table. The coefficient of static friction between the block and the table is . The coefficient of kinetic friction is , with . Part A m μs μk μk < μs If the block is at rest (and the only forces acting on the block are the force due to gravity and the normal force from the table), what is the magnitude of the force due to friction? You did not open hints for this part. ANSWER: Part B Suppose you want to move the block, but you want to push it with the least force possible to get it moving. With what force must you be pushing the block just before the block begins to move? Express the magnitude of in terms of some or all the variables , , and , as well as the acceleration due to gravity . You did not open hints for this part. ANSWER: Part C Suppose you push horizontally with half the force needed to just make the block move. What is the magnitude of the friction force? Express your answer in terms of some or all of the variables , , and , as well as the acceleration due to gravity . You did not open hints for this part. Ffriction = F F μs μk m g F = μs μk m g ANSWER: Part D Suppose you push horizontally with precisely enough force to make the block start to move, and you continue to apply the same amount of force even after it starts moving. Find the acceleration of the block after it begins to move. Express your answer in terms of some or all of the variables , , and , as well as the acceleration due to gravity . You did not open hints for this part. ANSWER: Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. Ffriction = a μs μk m g a =

Chapter 11 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, April 18, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Understanding Work and Kinetic Energy Learning Goal: To learn about the Work-Energy Theorem and its basic applications. In this problem, you will learn about the relationship between the work done on an object and the kinetic energy of that object. The kinetic energy of an object of mass moving at a speed is defined as . It seems reasonable to say that the speed of an object–and, therefore, its kinetic energy–can be changed by performing work on the object. In this problem, we will explore the mathematical relationship between the work done on an object and the change in the kinetic energy of that object. First, let us consider a sled of mass being pulled by a constant, horizontal force of magnitude along a rough, horizontal surface. The sled is speeding up. Part A How many forces are acting on the sled? ANSWER: Part B This question will be shown after you complete previous question(s). Part C K m v K = (1/2)mv2 m F one two three four This question will be shown after you complete previous question(s). Part D This question will be shown after you complete previous question(s). Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). Part G This question will be shown after you complete previous question(s). Part H This question will be shown after you complete previous question(s). Part I Typesetting math: 91% This question will be shown after you complete previous question(s). Part J This question will be shown after you complete previous question(s). Part K This question will be shown after you complete previous question(s). Work-Energy Theorem Reviewed Learning Goal: Review the work-energy theorem and apply it to a simple problem. If you push a particle of mass in the direction in which it is already moving, you expect the particle’s speed to increase. If you push with a constant force , then the particle will accelerate with acceleration (from Newton’s 2nd law). Part A Enter a one- or two-word answer that correctly completes the following statement. If the constant force is applied for a fixed interval of time , then the _____ of the particle will increase by an amount . You did not open hints for this part. ANSWER: M F a = F/M t at Typesetting math: 91% Part B Enter a one- or two-word answer that correctly completes the following statement. If the constant force is applied over a given distance , along the path of the particle, then the _____ of the particle will increase by . ANSWER: Part C If the initial kinetic energy of the particle is , and its final kinetic energy is , express in terms of and the work done on the particle. ANSWER: Part D In general, the work done by a force is written as . Now, consider whether the following statements are true or false: The dot product assures that the integrand is always nonnegative. The dot product indicates that only the component of the force perpendicular to the path contributes to the integral. The dot product indicates that only the component of the force parallel to the path contributes to the integral. Enter t for true or f for false for each statement. Separate your responses with commas (e.g., t,f,t). ANSWER: D FD Ki Kf Kf Ki W Kf = F W = ( ) d f i F r r Typesetting math: 91% Part E Assume that the particle has initial speed . Find its final kinetic energy in terms of , , , and . You did not open hints for this part. ANSWER: Part F What is the final speed of the particle? Express your answer in terms of and . ANSWER: ± The Work Done in Pulling a Supertanker Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 2.20×106 , one at an angle 10.0 west of north, and the other at an angle 10.0 east of north, as they pull the tanker a distance 0.660 toward the north. Part A What is the total work done by the two tugboats on the supertanker? Express your answer in joules, to three significant figures. vi Kf vi M F D Kf = Kf M vf = N km Typesetting math: 91% You did not open hints for this part. ANSWER: Energy Required to Lift a Heavy Box As you are trying to move a heavy box of mass , you realize that it is too heavy for you to lift by yourself. There is no one around to help, so you attach an ideal pulley to the box and a massless rope to the ceiling, which you wrap around the pulley. You pull up on the rope to lift the box. Use for the magnitude of the acceleration due to gravity and neglect friction forces. Part A Once you have pulled hard enough to start the box moving upward, what is the magnitude of the upward force you must apply to the rope to start raising the box with constant velocity? Express the magnitude of the force in terms of , the mass of the box. J m g F m Typesetting math: 91% You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Pulling a Block on an Incline with Friction A block of weight sits on an inclined plane as shown. A force of magnitude is applied to pull the block up the incline at constant speed. The coefficient of kinetic friction between the plane and the block is . Part A F = mg F μ Typesetting math: 91% What is the total work done on the block by the force of friction as the block moves a distance up the incline? Express the work done by friction in terms of any or all of the variables , , , , , and . You did not open hints for this part. ANSWER: Part B What is the total work done on the block by the applied force as the block moves a distance up the incline? Express your answer in terms of any or all of the variables , , , , , and . ANSWER: Now the applied force is changed so that instead of pulling the block up the incline, the force pulls the block down the incline at a constant speed. Wfric L μ m g L F Wfric = WF F L μ m g L F WF = Typesetting math: 91% Part C What is the total work done on the block by the force of friction as the block moves a distance down the incline? Express your answer in terms of any or all of the variables , , , , , and . ANSWER: Part D What is the total work done on the box by the appled force in this case? Express your answer in terms of any or all of the variables , , , , , and . ANSWER: When Push Comes to Shove Two forces, of magnitudes = 75.0 and = 25.0 , act in opposite directions on a block, which sits atop a frictionless surface, as shown in the figure. Initially, the center of the block is at position = -1.00 . At some later time, the block has moved to the right, and its center is at a new position, = 1.00 . Wfric L μ m g L F Wfric = WF μ m g L F WF = F1 N F2 N xi cm xf cm Typesetting math: 91% Part A Find the work done on the block by the force of magnitude = 75.0 as the block moves from = -1.00 to = 1.00 . Express your answer numerically, in joules. You did not open hints for this part. ANSWER: Part B Find the work done by the force of magnitude = 25.0 as the block moves from = -1.00 to = 1.00 . Express your answer numerically, in joules. You did not open hints for this part. ANSWER: W1 F1 N xi cm xf cm W1 = J W2 F2 N xi cm xf cm Typesetting math: 91% Part C What is the net work done on the block by the two forces? Express your answer numerically, in joules. ANSWER: Part D Determine the change in the kinetic energy of the block as it moves from = -1.00 to = 1.00 . Express your answer numerically, in joules. You did not open hints for this part. ANSWER: Work from a Constant Force Learning Goal: W2 = J Wnet Wnet = J Kf − Ki xi cm xf cm Kf − Ki = J Typesetting math: 91% To understand how to compute the work done by a constant force acting on a particle that moves in a straight line. In this problem, you will calculate the work done by a constant force. A force is considered constant if is independent of . This is the most frequently encountered situation in elementary Newtonian mechanics. Part A Consider a particle moving in a straight line from initial point B to final point A, acted upon by a constant force . The force (think of it as a field, having a magnitude and direction at every position ) is indicated by a series of identical vectors pointing to the left, parallel to the horizontal axis. The vectors are all identical only because the force is constant along the path. The magnitude of the force is , and the displacement vector from point B to point A is (of magnitude , making and angle (radians) with the positive x axis). Find , the work that the force performs on the particle as it moves from point B to point A. Express the work in terms of , , and . Remember to use radians, not degrees, for any angles that appear in your answer. You did not open hints for this part. ANSWER: Part B Now consider the same force acting on a particle that travels from point A to point B. The displacement vector now points in the opposite direction as it did in Part A. Find the work done by in this case. Express your answer in terms of , , and . F( r) r F r F L L WBA F L F WBA = F L WAB F Typesetting math: 91% L F You did not open hints for this part. ANSWER: ± Vector Dot Product Let vectors , , and . Calculate the following: Part A You did not open hints for this part. ANSWER: WAB = A = (2, 1,−4) B = (−3, 0, 1) C = (−1,−1, 2) Typesetting math: 91% Part B What is the angle between and ? Express your answer using one significant figure. You did not open hints for this part. ANSWER: Part C ANSWER: Part D ANSWER: A B = AB A B AB = radians 2B 3C = Typesetting math: 91% Part E Which of the following can be computed? You did not open hints for this part. ANSWER: and are different vectors with lengths and respectively. Find the following: Part F Express your answer in terms of You did not open hints for this part. ANSWER: 2(B 3C) = A B C A (B C) A (B + C) 3 A V 1 V 2 V1 V2 V1 Typesetting math: 91% Part G If and are perpendicular, You did not open hints for this part. ANSWER: Part H If and are parallel, Express your answer in terms of and . You did not open hints for this part. ANSWER: ± Tactics Box 11.1 Calculating the Work Done by a Constant Force V = 1 V 1 V 1 V 2 V = 1 V 2 V 1 V 2 V1 V2 V = 1 V 2 Typesetting math: 91% Learning Goal: To practice Tactics Box 11.1 Calculating the Work Done by a Constant Force. Recall that the work done by a constant force at an angle to the displacement is . The vector magnitudes and are always positive, so the sign of is determined entirely by the angle between the force and the displacement. W F d W = Fd cos F d W Typesetting math: 91% TACTICS BOX 11.1 Calculating the work done by a constant force Force and displacement Work Sign of Energy transfer Energy is transferred into the system. The particle speeds up. increases. No energy is transferred. Speed and are constant. Energy is transferred out of the system. The particle slows down. decreases. A box has weight of magnitude = 2.00 accelerates down a rough plane that is inclined at an angle = 30.0 above the horizontal, as shown at left. The normal force acting on the box has a magnitude = 1.732 , the coefficient of kinetic friction between the box and the plane is = 0.300, and the displacement of the box is 1.80 down the inclined plane. W W 0 F(“r) + K < 90 F("r) cos + 90 0 0 K > 90 F(“r) cos − K 180 −F(“r) − FG N n N μk d m Typesetting math: 91% Part A What is the work done on the box by gravity? Express your answers in joules to two significant figures. You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Wgrav Wgrav = J Typesetting math: 91% Work and Potential Energy on a Sliding Block with Friction A block of weight sits on a plane inclined at an angle as shown. The coefficient of kinetic friction between the plane and the block is . A force is applied to push the block up the incline at constant speed. Part A What is the work done on the block by the force of friction as the block moves a distance up the incline? Express your answer in terms of some or all of the following: , , , . You did not open hints for this part. ANSWER: w μ F Wf L μ w L Wf = Typesetting math: 91% Part B What is the work done by the applied force of magnitude ? Express your answer in terms of some or all of the following: , , , . ANSWER: Part C What is the change in the potential energy of the block, , after it has been pushed a distance up the incline? Express your answer in terms of some or all of the following: , , , . ANSWER: Part D This question will be shown after you complete previous question(s). Part E This question will be shown after you complete previous question(s). W F μ w L W = “U L μ w L “U = Typesetting math: 91% Part F This question will be shown after you complete previous question(s). Where’s the Energy? Learning Goal: To understand how to apply the law of conservation of energy to situations with and without nonconservative forces acting. The law of conservation of energy states the following: In an isolated system the total energy remains constant. If the objects within the system interact through gravitational and elastic forces only, then the total mechanical energy is conserved. The mechanical energy of a system is defined as the sum of kinetic energy and potential energy . For such systems where no forces other than the gravitational and elastic forces do work, the law of conservation of energy can be written as , where the quantities with subscript “i” refer to the “initial” moment and those with subscript “f” refer to the final moment. A wise choice of initial and final moments, which is not always obvious, may significantly simplify the solution. The kinetic energy of an object that has mass \texttip{m}{m} and velocity \texttip{v}{v} is given by \large{K=\frac{1}{2}mv^2}. Potential energy, instead, has many forms. The two forms that you will be dealing with most often in this chapter are the gravitational and elastic potential energy. Gravitational potential energy is the energy possessed by elevated objects. For small heights, it can be found as U_{\rm g}=mgh, where \texttip{m}{m} is the mass of the object, \texttip{g}{g} is the acceleration due to gravity, and \texttip{h}{h} is the elevation of the object above the zero level. The zero level is the elevation at which the gravitational potential energy is assumed to be (you guessed it) zero. The choice of the zero level is dictated by convenience; typically (but not necessarily), it is selected to coincide with the lowest position of the object during the motion explored in the problem. Elastic potential energy is associated with stretched or compressed elastic objects such as springs. For a spring with a force constant \texttip{k}{k}, stretched or compressed a distance \texttip{x}{x}, the associated elastic potential energy is \large{U_{\rm e}=\frac{1}{2}kx^2}. When all three types of energy change, the law of conservation of energy for an object of mass \texttip{m}{m} can be written as K U Ki + Ui = Kf + Uf Typesetting math: 91% \large{\frac{1}{2}mv_{\rm i}^2+mgh_{\rm i}+\frac{1}{2}kx_{\rm i}^2=\frac{1}{2}mv_{\rm f \hspace{1 pt}}^2+mgh_{\rm f \hspace{1 pt}}+\frac{1}{2}kx_{\rm f \hspace{1 pt}}^2}. The gravitational force and the elastic force are two examples of conservative forces. What if nonconservative forces, such as friction, also act within the system? In that case, the total mechanical energy would change. The law of conservation of energy is then written as \large{\frac{1}{2}mv_{\rm i}^2+mgh_{\rm i}+\frac{1}{2}kx_{\rm i}^2+W_{\rm nc}=\frac{1}{2}mv_{\rm f \hspace{1 pt}}^2+mgh_{\rm f \hspace{1 pt}}+\frac{1}{2}kx_{\rm f \hspace{1 pt}}^2}, where \texttip{W_{\rm nc}}{W_nc} represents the work done by the nonconservative forces acting on the object between the initial and the final moments. The work \texttip{W_{\rm nc}}{W_nc} is usually negative; that is, the nonconservative forces tend to decrease, or dissipate, the mechanical energy of the system. In this problem, we will consider the following situation as depicted in the diagram : A block of mass \texttip{m}{m} slides at a speed \texttip{v}{v} along a horizontal, smooth table. It next slides down a smooth ramp, descending a height \texttip{h}{h}, and then slides along a horizontal rough floor, stopping eventually. Assume that the block slides slowly enough so that it does not lose contact with the supporting surfaces (table, ramp, or floor). You will analyze the motion of the block at different moments using the law of conservation of energy. Part A Which word in the statement of this problem allows you to assume that the table is frictionless? ANSWER: Part B straight smooth horizontal Typesetting math: 91% This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Part D This question will be shown after you complete previous question(s). Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). Part G This question will be shown after you complete previous question(s). Part H Typesetting math: 91% This question will be shown after you complete previous question(s). Part I This question will be shown after you complete previous question(s). Part J This question will be shown after you complete previous question(s). Part K This question will be shown after you complete previous question(s). Sliding In Socks Suppose that the coefficient of kinetic friction between Zak’s feet and the floor, while wearing socks, is 0.250. Knowing this, Zak decides to get a running start and then slide across the floor. Part A If Zak’s speed is 3.00 \rm m/s when he starts to slide, what distance \texttip{d}{d} will he slide before stopping? Express your answer in meters. ANSWER: Typesetting math: 91% Part B This question will be shown after you complete previous question(s). Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. \rm m Typesetting math: 91%

Two force vectors F1 and F2 are applied at the origin of the system of coordinates, point O of coordinates (0,0,0). The two force are expressed in Newton units (N). The vector expressed of the first fore is F1=(-120i+60j+40k) N. The second force F2 has a magnitude of 85 N and its direction is defined by the line between point O and B (4, -3, 5), where these coordinates are in meters (m). (a) Find unit vector ef1alone F1. (b) Find unit vector ef2alone F2. (c) find the angle O between F1 and F2 using the dot product operation. (d) Find the force vector resultant R=F1+F2. (e) find the direction cosine, cosOxof vector R.

Chapter 3 Practice Problems (Practice – no credit) Due: 11:59pm on Wednesday, February 12, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Tactics Box 3.1 Determining the Components of a Vector Learning Goal: To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector is decomposed into component vectors and parallel to the coordinate axes, we can describe each component vector with a single number (a scalar) called the component. This tactics box describes how to determine the x component and y component of vector , denoted and . TACTICS BOX 3.1 Determining the components of a vector The absolute value of the x component is the magnitude of the 1. component vector . 2. The sign of is positive if points in the positive x direction; it is negative if points in the negative x direction. 3. The y component is determined similarly. Part A What is the magnitude of the component vector shown in the figure? Express your answer in meters to one significant figure. A A x A y A Ax Ay |Ax| Ax A x Ax A x A x Ay A x

To identify the correct notation for a point and a vector, determine the position vector of a point relative to another point , and calculate the corresponding unit vector. although vectors are often constructed from points , points are not vectors, vectors are commonly constructed from either : (1) the origin to a point or (2) a starting point to an ending point. Part A ) as shown on the coordinate system, points A and B have the following distance from the origin : xA = 2.70 ft , zA = 2.50 ft , xB = 1.10 ft , and zB = 1.70 ft. which of the following statements correctly describes the location and position vector of point A from the origin ? Use appropriate notation for the location and position vector of a point,

Chapter 5 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, March 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Tactics Box 5.1 Drawing Force Vectors Learning Goal: To practice Tactics Box 5.1 Drawing Force Vectors. To visualize how forces are exerted on objects, we can use simple diagrams such as vectors. This Tactics Box illustrates the process of drawing a force vector by using the particle model, in which objects are treated as points. TACTICS BOX 5.1 Drawing force vectors Represent the object 1. as a particle. 2. Place the tail of the force vector on the particle. 3. Draw the force vector as an arrow pointing in the proper direction and with a length proportional to the size of the force. 4. Give the vector an appropriate label. The resulting diagram for a force exerted on an object is shown in the drawing. Note that the object is represented as a black dot. Part A A book lies on a table. A pushing force parallel to the table top and directed to the right is exerted on the book. Follow the steps above to draw the force vector . Use the black dot as the particle representing the book. F F push F push

Lab 1: Introduction to Motion  You must make the following changes to your lab manual before coming to lab, not during lab!  Do not plan to consult this sheet during lab. There is not enough time.  The required changes must be in your lab manual in the proper sequence to complete the lab in a smooth and timely manner.  You should bring this paper to lab but only for reference to the images printed below. You have been warned! A note about vector addition: Adding Vectors: To add these two vectors: means to place them head-to-tail like so: and therefore they equal: Subtracting Vectors: Subtracting these two vectors: is the same as the sum of one vector and the negative of the other: which is the same as: which means to place them head-to-tail like so: and therefore they equal: Pg. 7 Activity 1-3 Cross off Step 1 Cross off Step 2 Pg. 7 Step 3) Replace “Try to make each of the graphs …” with “Try to make one of the graphs…” Pg. 7 Step 4) Replace this step with: “Describe how you must move to produce the graph you selected. Note if you selected graph C your description is at the top of page 8. Pg. 8 Activity 2-1 Step 2) Replace: “(Just draw smooth patterns; leave out…” with “(Quickly draw smooth patterns; leave out…” Then highlight this entire sentence. + − + (− ) + Pg. 10 Step 3) Where it states “(Be sure to adjust the time scale to 15 s.)” The way to do this is to click this clock icon And change the “Duration:” value Pg. 11 Question 2-3) At the end of the question add the following: “See the top of page 12 for the rest of the question.” Pg. 13 Step 2) Highlight the part that states: “Get the times right. Get the velocities right. Each person should take a turn.” At the end of the paragraph add: “But do not spend too much time getting things perfect.” Pg. 15 Step 1) Where is states: “Use the analysis feature of the software to read values of velocity…” Do this: Click here and then move the mouse over the graph. You can now quickly read data from the graph.

Extra Credit Due: 11:59pm on Thursday, May 15, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A Man Running to Catch a Bus A man is running at speed (much less than the speed of light) to catch a bus already at a stop. At , when he is a distance from the door to the bus, the bus starts moving with the positive acceleration . Use a coordinate system with at the door of the stopped bus. Part A What is , the position of the man as a function of time? Answer symbolically in terms of the variables , , and . Hint 1. Which equation should you use for the man’s speed? Because the man’s speed is constant, you may use . ANSWER: Correct Part B What is , the position of the bus as a function of time? Answer symbolically in terms of and . = Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 1 of 57 5/9/2014 8:02 PM Hint 1. Which equation should you use for the bus’s acceleration? Because the bus has constant acceleration, you may use . Recall that . ANSWER: Correct Part C What condition is necessary for the man to catch the bus? Assume he catches it at time . Hint 1. How to approach this problem If the man is to catch the bus, then at some moment in time , the man must arrive at the position of the door of the bus. How would you express this condition mathematically? ANSWER: Correct Part D Inserting the formulas you found for and into the condition , you obtain the following: , or . Intuitively, the man will not catch the bus unless he is running fast enough. In mathematical terms, there is a constraint on the man’s speed so that the equation above gives a solution for that is a real positive number. Find , the minimum value of for which the man will catch the bus. Express the minimum value for the man’s speed in terms of and . = Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 2 of 57 5/9/2014 8:02 PM Hint 1. Consider the discriminant Use the quadratic equation to solve: . What is the discriminant (the part under the radical) of the solution for ? Hint 1. The quadratic formula Recall: If then ANSWER: Hint 2. What is the constraint? To get a real value for , the discriminant must be greater then or equal to zero. This condition yields a constraint that exceed . ANSWER: Correct Part E Assume that the man misses getting aboard when he first meets up with the bus. Does he get a second chance if he continues to run at the constant speed ? = = Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 3 of 57 5/9/2014 8:02 PM Hint 1. What is the general quadratic equation? The general quadratic equation is , where , , and are constants. Depending on the value of the discriminant, , the equation may have two real valued 1. solutions if , 2. one real valued solution if , or 3. two complex valued solutions if . In this case, every real valued solution corresponds to a time at which the man is at the same position as the door of the bus. ANSWER: Correct Adding and Subtracting Vectors Conceptual Question Six vectors (A to F) have the magnitudes and directions indicated in the figure. Part A Which two vectors, when added, will have the largest (positive) x component? Hint 1. Largest x component The two vectors with the largest x components will, when combined, give the resultant with the largest x component. Keep in mind that positive x components are larger than negative x components. No; there is no chance he is going to get aboard. Yes; he will get a second chance Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 4 of 57 5/9/2014 8:02 PM ANSWER: Correct Part B Which two vectors, when added, will have the largest (positive) y component? Hint 1. Largest y component The two vectors with the largest y components will, when combined, give the resultant with the largest y component. Keep in mind that positive y components are larger than negative y components. ANSWER: Correct Part C Which two vectors, when subtracted (i.e., when one vector is subtracted from the other), will have the largest magnitude? Hint 1. Subtracting vectors To subtract two vectors, add a vector with the same magnitude but opposite direction of one of the vectors to the other vector. ANSWER: C and E E and F A and F C and D B and D C and D A and F E and F A and B E and D Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 5 of 57 5/9/2014 8:02 PM Correct Tactics Box 3.1 Determining the Components of a Vector Learning Goal: To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector is decomposed into component vectors and parallel to the coordinate axes, we can describe each component vector with a single number (a scalar) called the component. This tactics box describes how to determine the x component and y component of vector , denoted and . TACTICS BOX 3.1 Determining the components of a vector The absolute value of the x component is the magnitude of the 1. component vector . The sign of is positive if points in the positive x direction; it is negative if points in the negative x direction. 2. 3. The y component is determined similarly. Part A What is the magnitude of the component vector shown in the figure? Express your answer in meters to one significant figure. ANSWER: A and F A and E D and B C and D E and F Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 6 of 57 5/9/2014 8:02 PM Correct Part B What is the sign of the y component of vector shown in the figure? ANSWER: Correct Part C Now, combine the information given in the tactics box above to find the x and y components, and , of vector shown in the figure. Express your answers, separated by a comma, in meters to one significant figure. = 5 positive negative Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 7 of 57 5/9/2014 8:02 PM ANSWER: Correct Conceptual Problem about Projectile Motion Learning Goal: To understand projectile motion by considering horizontal constant velocity motion and vertical constant acceleration motion independently. Projectile motion refers to the motion of unpowered objects (called projectiles) such as balls or stones moving near the surface of the earth under the influence of the earth’s gravity alone. In this analysis we assume that air resistance can be neglected. An object undergoing projectile motion near the surface of the earth obeys the following rules: An object undergoing projectile motion travels horizontally at a constant rate. That is, the x component of its velocity, , is constant. 1. An object undergoing projectile motion moves vertically with a constant downward acceleration whose magnitude, denoted by , is equal to 9.80 near the surface of the earth. Hence, the y component of its velocity, , changes continuously. 2. An object undergoing projectile motion will undergo the horizontal and vertical motions described above from the instant it is launched until the instant it strikes the ground again. Even though the horizontal and vertical motions can be treated independently, they are related by the fact that they occur for exactly the same amount of time, namely the time the projectile is in the air. 3. The figure shows the trajectory (i.e., the path) of a ball undergoing projectile motion over level ground. The time corresponds to the moment just after the ball is launched from position and . Its launch velocity, also called the initial velocity, is . Two other points along the trajectory are indicated in the figure. One is the moment the ball reaches the peak of its trajectory, at time with velocity . Its position at this moment is denoted by or since it is at its maximum height. The other point, at time with velocity , corresponds to the moment just before the ball strikes the ground on the way back down. At this time its position is , also known as ( since it is at its maximum horizontal range. Projectile motion is symmetric about the peak, provided the object lands at the same vertical height from which is was launched, as is the case here. Hence . Part A , = -2,-5 , Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 8 of 57 5/9/2014 8:02 PM How do the speeds , , and (at times ,